Gergonne point Harmonic quadrilateral
by niwobin, May 17, 2025, 8:17 PM
Attachments:
Functional Equation from IMO
by prtoi, May 17, 2025, 8:07 PM
Question: 
Solve for f:Z-->Z
My solution:
At a=0,
Take t=f(b) to get
Therefore, f(x)=2x+n where n=f(0)
Could someone please clarify if this is right or wrong?
Solve for f:Z-->Z
My solution:
At a=0,
Take t=f(b) to get
Therefore, f(x)=2x+n where n=f(0)
Could someone please clarify if this is right or wrong?
Combi that will make you question every choice in your life so far
by blug, May 17, 2025, 5:46 PM
and 
are standing in front of the room in which there is 
. They know that there is a chessboard in the room and that on every square there is a coin. Every coin is black on one side and white on the other side and is flipped randomly. enters the room and then points at exactly one square on the chessboard. After that, must flip exactly one coin of his choice on the chessboard to the other side and leave. Finally, enters the room ( and haven't met again after entered the room) and he has to guess which square did point at.What strategy do
and have that will make this happen every time?can you solve this..?
by Jackson0423, May 8, 2025, 4:17 PM
Find the number of integer pairs 
satisfying the equation
![\[ 4x^2 - 3y^2 = 1 \]](//latex.artofproblemsolving.com/d/2/8/d284c2c3a7d7fbe12a7988a4534df16e7adc8947.png)
such that 
.
satisfying the equation![\[ 4x^2 - 3y^2 = 1 \]](http://latex.artofproblemsolving.com/d/2/8/d284c2c3a7d7fbe12a7988a4534df16e7adc8947.png)
such that 
.Concurrence of lines defined by intersections of circles
by Lukaluce, Apr 14, 2025, 10:57 AM
Let 
be an acute-angled triangle and 
, and 
be the feet of the altitudes from 
, and 
, respectively. On the rays 
, and 
, we have points 
, and 
respectively, lying outside of , such that
![\[\frac{A_1A_2}{AA_1} = \frac{B_1B_2}{BB_1} = \frac{C_1C_2}{CC_1}.\]](//latex.artofproblemsolving.com/d/f/a/dfa94e10638971d53066c20c4589a9c125059911.png)
If the intersections of 
and 
, 
and 
, and 
and 
are 
, and 
respectively, prove that 
, and 
have a common point.
be an acute-angled triangle and 
, and 
be the feet of the altitudes from 
, and , respectively. On the rays 
, and 
, we have points 
, and 
respectively, lying outside of , such that![\[\frac{A_1A_2}{AA_1} = \frac{B_1B_2}{BB_1} = \frac{C_1C_2}{CC_1}.\]](http://latex.artofproblemsolving.com/d/f/a/dfa94e10638971d53066c20c4589a9c125059911.png)
If the intersections of 
and 
, 
and 
, and 
and 
are 
, and 
respectively, prove that 
, and 
have a common point.Factorial Divisibility
by Aryan-23, Jul 9, 2023, 5:02 AM
Find all positive integers 
such that
such that
This post has been edited 1 time. Last edited by Aryan-23, Aug 7, 2023, 7:29 AM
Functional equation
by Pmshw, May 8, 2022, 3:57 PM
Find all functions 
such that for any real value of 
we have:
such that for any real value of 
we have:
This post has been edited 1 time. Last edited by Pmshw, May 8, 2022, 3:58 PM
Hard Geometry
by Jalil_Huseynov, Dec 26, 2021, 7:10 PM
Let triangle 
be a triangle with incenter 
and circumcircle 
with circumcenter 
. The incircle touches 
at 
respectively. 
is another intersection point of external bisector of 
with , and 
is 
incircle touch point to . Let 
be points lie on . 
intersect 
at 
. Assume that 
. Suppose that 
and 
are concyclic, and 
are concurrent.
Prove that
are concurrent.
, tangent line to at and 
are concurrent.
Proporsed by wassupevery1 and k12byda5h
be a triangle with incenter 
and circumcircle 
with circumcenter 
. The incircle touches 
at 
respectively. 
is another intersection point of external bisector of 
with , and 
is 
incircle touch point to . Let 
be points lie on . 
intersect 
at 
. Assume that 
. Suppose that 
and 
are concyclic, and 
are concurrent.Prove that
are concurrent. 
, tangent line to at and 
are concurrent.Proporsed by wassupevery1 and k12byda5h
This post has been edited 1 time. Last edited by Jalil_Huseynov, Dec 28, 2021, 12:36 PM
Algebra form IMO Shortlist
by Abbas11235, Jul 10, 2018, 11:54 AM
Let 
be a real number. Gugu has a napkin with ten distinct real numbers written on it, and he writes the following three lines of real numbers on the blackboard:
such that, regardless of the numbers on Gugu's napkin, every number in the second line is also a number in the third line.
be a real number. Gugu has a napkin with ten distinct real numbers written on it, and he writes the following three lines of real numbers on the blackboard:
- In the first line, Gugu writes down every number of the form
, where 
and 
are two (not necessarily distinct) numbers on his napkin. - In the second line, Gugu writes down every number of the form
, where and are
two (not necessarily distinct) numbers from the first line. - In the third line, Gugu writes down every number of the form
, where 
are four (not necessarily distinct) numbers from the first line.
,
and 
are two (not necessarily distinct) numbers on his napkin.
,
,
are four (not necessarily distinct) numbers from the first line. such that, regardless of the numbers on Gugu's napkin, every number in the second line is also a number in the third line.This post has been edited 13 times. Last edited by levans, Aug 19, 2018, 6:27 PM
IMO 2011 Problem 6
by liberator, Jul 21, 2015, 4:57 PM
Problem: Let be an acute triangle with circumcircle 
. Let 
be a tangent line to , and let 
and 
be the lines obtained by reflecting in the lines 
, 
and 
, respectively. Show that the circumcircle of the triangle determined by the lines and is tangent to the circle .
Proposed by Japan
be an acute triangle with circumcircle 
. Let 
be a tangent line to , and let 
and 
be the lines obtained by reflecting in the lines 
, 
and 
, respectively. Show that the circumcircle of the triangle determined by the lines and is tangent to the circle .Proposed by Japan
![[asy]
import graph; size(10cm); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(9); pointfontpen=fp; real xmin=2.5,xmax=14,ymin=-2,ymax=8.8;
pen wwwwww=rgb(0.4,0.4,0.4), wwzzcc=rgb(0.4,0.6,0.8), yqqqyq=rgb(0.5019607843137255,0.,0.5019607843137255), qqcccc=rgb(0,0.8,0.8), cqcqcq=rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882);
pair A=(9.,3.2), B=(7.6,1.), C=(11.06,1.), P=(10.9037267872876,2.2894742445215552), A_1=(4.225302839837132,4.713218228557546), B_1=(13.318947812589142,4.209009023926156), C_1=(11.23751359726076,0.33222222307412996), M=(9.699442104579619,-0.3030337832920753), P_a=(10.9037267872876,-0.28947424452155524), P_b=(9.783568496159925,1.2405987537383683), P_c=(7.368886499833012,4.538918063810838), H=(9.,2.310909090909089), I_1=(10.610893030990937,2.689468626599);
D(A--B--C--cycle,wwwwww+linewidth(1)); D(A_1--B_1--C_1--cycle,wwzzcc+linewidth(1)); D(A--A_1,linetype("4 4")+wwzzcc);
D(CR((9.33,1.4445454545454548),1.786202861143444),green); D((xmin,-1.8625555265220795*xmin+22.598270831870913)--(xmax,-1.8625555265220795*xmax+22.598270831870913),wwzzcc); D(CR((8.755537154895718,4.161938324953828),4.563653413817887),yqqqyq); D(CR((9.33,0.5554545454545452),1.786202861143444),red); D(CR((10.73,2.7554545454545445),1.786202861143444),red); D(CR((7.27,2.7554545454545463),1.786202861143444),red); D(P_a--P_c,linetype("4 4")+wwzzcc); D(CR((5.579166076673129,0.695625707960863),4.239574899162703),linetype("0 2")+qqcccc+linewidth(1)); D(CR((10.26502172862502,2.9510482106080516),3.3028669264989134),linetype("0 2")+qqcccc+linewidth(1)); D(CR((10.293309185404135,0.43870973889632964),0.9501902769422019),linetype("0 2")+qqcccc+linewidth(1)); D(M--A_1,wwzzcc); D(M--B_1,wwzzcc);
D(A); MP("A",(9.078267017098327,3.3218831364461052),NE*lsf); D(B); MP("B",(7.208843909967847,0.5368242217415146),NE*lsf); D(C); MP("C",(11.233826314095717,0.7657331736350426),NE*lsf); D(P); MP("P",(10.985841616211061,2.406247328871993),NE*lsf); D(C_1); MP("C_1",(11.367356536033608,0.11715780993671321),NE*lsf); D(A_1); MP("A_1",(3.698906647600416,4.981473037674183),NE*lsf); D(B_1); MP("B_1",(13.484764341048743,4.046761484108944),NE*lsf); D(M); MP("M",(9.688690888814401,-0.7030992676817621),NE*lsf); D(P_a); MP("P_a",(10.966765870219934,-0.5886447917349982),NE*lsf); D(P_b); MP("P_b",(9.898524094716802,1.1663238394487165),NE*lsf); D(P_c); MP("P_c",(7.11346518001221,4.752564085780656),NE*lsf); D(H); MP("H",(9.078267017098327,2.4253230748631207),NE*lsf); D(I_1); MP("I_1",(10.680629680353023,2.8068379946856674),NE*lsf);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* Labels, by liberator */
defaultpen(fontsize(9pt));
label("$\Gamma_a$",(10,-1.2),S*lsf);
label("$\Gamma_b$",(12,4),dir(30)*lsf);
label("$\Gamma_c$",(6,1),N*lsf);
label("$\Gamma_1$",(5,7),N*lsf);
[/asy]](http://latex.artofproblemsolving.com/f/b/8/fb89592e2b7f740039a8d16e7184e085a07a93d6.png)
Let 
,
analogously; let 
be the circumcircle of 
.
,
.
is tangent to 
,
,
is the Steiner line of 
w.r.t 
.
lie on 
concur at 
be the incenter of 
are the centers of 
then 
.
;
.
:
.
,
is the internal bisector of 
.
are the internal bisectors of 
respectively. Hence 
concur at 
as the Miquel point of the line 
.
is a 
![[asy]
import graph; size(10cm); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(9); pointfontpen=fp; real xmin=2.5,xmax=14,ymin=-2,ymax=8.8;
pen wwwwww=rgb(0.4,0.4,0.4), wwzzcc=rgb(0.4,0.6,0.8), yqqqyq=rgb(0.5019607843137255,0.,0.5019607843137255), cqcqcq=rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882);
pair A=(9.,3.2), B=(7.6,1.), C=(11.06,1.), P=(10.9037267872876,2.2894742445215552), A_1=(4.225302839837132,4.713218228557546), B_1=(13.318947812589142,4.209009023926156), C_1=(11.23751359726076,0.33222222307412996), M=(9.699442104579619,-0.3030337832920753), P_a=(10.9037267872876,-0.28947424452155524), P_b=(9.783568496159925,1.2405987537383683), P_c=(7.368886499833012,4.538918063810838), H=(9.,2.310909090909089), I_1=(10.610893030990937,2.689468626599), A_2=(7.556877292936604,1.6603150697678517), B_2=(11.116107846985981,1.4629688117894197), K=(12.028150798496817,7.3426508899567695);
D(A--B--C--cycle,wwwwww+linewidth(1)); D(A_1--B_1--C_1--cycle,wwzzcc+linewidth(1)); D(I_1--A_2--B_2--cycle,cqcqcq); D(A_1--K--B_1--cycle,wwzzcc); D(A--A_1,linetype("4 4")+wwzzcc);
D(CR((9.33,1.4445454545454548),1.786202861143444),green); D((xmin,-1.8625555265220795*xmin+22.598270831870913)--(xmax,-1.8625555265220795*xmax+22.598270831870913),wwzzcc); D(CR((8.755537154895718,4.161938324953828),4.563653413817887),yqqqyq); D(P_a--P_c,linetype("4 4")+wwzzcc); D(CR((5.579166076673129,0.695625707960863),4.239574899162703),linetype("0 2")+red+linewidth(1)); D(CR((10.26502172862502,2.9510482106080516),3.3028669264989134),linetype("0 2")+red+linewidth(1)); D(CR((10.293309185404135,0.43870973889632964),0.9501902769422019),linetype("0 2")+red+linewidth(1)); D(M--A_1,wwzzcc); D(M--B_1,wwzzcc); D((xmin,0.33698371098991503*xmin+3.289359997533001)--(xmax,0.33698371098991503*xmax+3.289359997533001),wwzzcc); D((xmin,-2.4276798224809775*xmin+36.54314988522595)--(xmax,-2.4276798224809775*xmax+36.54314988522595),wwzzcc); D(I_1--A_2--B_2--cycle,wwzzcc);
D(A); MP("A",(9.078267017098327,3.3218831364461052),NE*lsf); D(B); MP("B",(7.208843909967847,0.5368242217415146),NE*lsf); D(C); MP("C",(11.233826314095717,0.7657331736350426),NE*lsf); D(P); MP("P",(10.985841616211061,2.406247328871993),NE*lsf); D(C_1); MP("C_1",(11.367356536033608,0.11715780993671321),NE*lsf); D(A_1); MP("A_1",(3.698906647600416,4.981473037674183),NE*lsf); D(B_1); MP("B_1",(13.484764341048743,4.046761484108944),NE*lsf); D(M); MP("M",(9.688690888814401,-0.7030992676817621),NE*lsf); D(P_a); MP("P_a",(10.966765870219934,-0.5886447917349982),NE*lsf); D(P_b); MP("P_b",(9.898524094716802,1.1663238394487165),NE*lsf); D(P_c); MP("P_c",(7.11346518001221,4.752564085780656),NE*lsf); D(H); MP("H",(9.078267017098327,2.4253230748631207),NE*lsf); D(I_1); MP("I_1",(10.680629680353023,2.8068379946856674),NE*lsf); D(A_2); MP("A_2",(7.266071147941229,1.9484294250849374),NE*lsf); D(B_2); MP("B_2",(11.386432282024735,1.5096872672890087),NE*lsf); D(K); MP("K",(12.07315913770532,7.5376230004852465),NE*lsf);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* Labels, by liberator */
defaultpen(fontsize(9pt));
label("$\Gamma_1$",(5,7),N*lsf);
[/asy]](http://latex.artofproblemsolving.com/3/0/c/30cdce277f8b9fc56322d3d0a0baa20ca56c49e3.png)
Suppose the tangents to 
at 
.
;
.
,
.
.
,
are collinear.
,
and 
are homothetic.
By a converse of Reim's theorem, 
are tangent at This post has been edited 2 times. Last edited by liberator, Jul 22, 2015, 1:35 PM
Multiple of multinomial coefficient is an integer
by orl, Mar 7, 2009, 8:00 PM
For 
, 
, and 
, let 
denote the greatest common divisor of 
.
Prove that
is an integer.
Dan Schwarz, Romania
, 
, and 
, let 
denote the greatest common divisor of 
.Prove that
is an integer.Dan Schwarz, Romania
October 2018
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